Consider $f : \Bbb R^2 \to \Bbb R$ defined by $f(x,y) = x + y$. For each of the following statements, state whether it is true
or false.

1. Image under $f$ of any open set if open,

2. Image under $f$ of any closed set is closed,

3. Image under $f$ of any dense set is dense,

4. Image under $f$ of any discrete set is discrete.

Every continuous subjective (onto) map between the metric spaces preserves the dense sets. The proof is essentially the same as above. Note that, in this problem given $f$ is a surjective continuous map.

Click 1. Image under $f$ of any open set if open,

2. Image under $f$ of any closed set is closed,

3. Image under $f$ of any dense set is dense,

4. Image under $f$ of any discrete set is discrete.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****Solution:****Option 1**: (**True**) Consider the function $f$. We claim that $f$ is an open map. Let $U$ be an open set in $\Bbb R^2$. We need to show that $f(U)$ is open in $\Bbb R$. Let $a \in f(U)$, we will show that there is a neighborhood of $a$ which is entirely contained in $f(U)$. Let $(x,y) \in U \subseteq \Bbb R^2$ be such that $f(x,y)=x+y = a$. Since $a \in f(U)$ this is possible. Now $(x,y) \in U$ and $U$ is open and so we can find an $\epsilon > 0$ such that the open set $V = (x-\epsilon,x+\epsilon) \times (y-\epsilon,y+\epsilon) \subseteq U$. Now, $f(V) = ((x+y)-2\epsilon, (x+y)+2\epsilon) = (a-2 \epsilon, a+2\epsilon)$ is an open interval. Since $V \subset U$ we have $f(V) \subseteq f(U)$ is an interval containing $a$.**Therefore $f(U)$ is open.****option 2:(False)**We will construct a closed set $F$ such that $f(F)$ is not closed. Let $F = \Bbb Z \times \sqrt r \Bbb Z$ where $r$ is not a perfect square integer. Then $f(F) = \Bbb Z + \sqrt r \Bbb Z$ which is a dense subset of $\Bbb R$. (**Why this is dense? please check my next post**). Therefore $f(F)$ is not closed.**option 4**:(**False**) Consider the example given in option 2. $F$ is discrete but $f(F)$ is not.**option 3**: (**True**) Let $D$ be a subset of a metric space $X$, then $D$ is dense in $X$ if given $x$ in $X$ there exists a sequence $d_n$ in $D$ such that $d_n$ converges to $x$. We will use this definition. Consider the function $f$ and let $D$ be a dense subset of $\Bbb R^2$. We claim that $f(D)$ is dense in $\Bbb R$. Let $a \in f(D)$ then there exists a point $(x,y) \in D$ such that $f(x,y) = a$. Now $(x,y) \in D$ and $D$ is dense. So there exists a sequence $(x_n,y_n) \in D$ such that $(x_n,y_n)$ converges to $(x,y)$. Since the function $f$ is continuous, we have $f((x_n,y_n)) = (x_n+y_n)$ converges to $x+y = a$. Since $(x_n,y_n)$ is a sequence in $D$ we have $(x_n+y_n)$ is a sequence in $f(D)$ which converges to $a$.**Therefore $f(D)$ is dense in $\Bbb R$.****THERE IS A MORE GENERAL RESULT IS TRUE:**Every continuous subjective (onto) map between the metric spaces preserves the dense sets. The proof is essentially the same as above. Note that, in this problem given $f$ is a surjective continuous map.

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I have doubt Sir

ReplyDeleteinverse image of open set is open.under continuous function

Yes.

DeleteNeed proof?

DeleteWhat is the doubt?

ReplyDelete